Question

if x2+x+1=0 then find the value of x2015+x2016

Answer 1

Since $x^{2}+x+1=0$

Consider $(x^{3}-1)$

Using the identity $(a^{3}-b^{3})=(a-b)(a^{2}+ab+b^{2})$

$(x^{3}-1)=(x-1)(x^{2}+x+1)$

Now, $\frac{(x^{3}-1)}{x-1}=(x^{2}+x+1)$

Since, $(x^{2}+x+1)=0$

$\frac{(x^{3}-1)}{x-1}=0$

${(x^{3}-1)}=0$

${x^{3}}=1$

so, x= 1

Now, we will find the value of $x^{2015}+x^{2016}$

= $1^{2015}+1^{2016}$

= 1 + 1

= 2

So, the value of the given expression is 2.

Answer 2

Answer:

$x^{2015}+x^{2016}$= $2$

Step-by-step explanation:

We are given $x^{2}+x+1=0$,

Using the identity, $(a^{3}-b^{3})=(a-b)(a^{2}+ab+b^{2})$

Let a= x and b=1 in the above identity,

⇒$(x^{3}-1)=(x-1)(x^{2}+x+1)$

⇒$\frac{(x^{3}-1)}{x-1}=x^{2}+x+1$

We know that  $x^{2}+x+1=0$, therefore

$\frac{(x^{3}-1)}{x-1}=0$

⇒$x^{3}-1=0$

⇒$x^{3}=1$

⇒$x=1$

Now, putting x=1 in equation, $x^{2015}+x^{2016}$

⇒$(1)^{2015}+(1)^{2016}$

⇒$1+1=2$

Therefore, the value of $x^{2015}+x^{2016}$= $2$